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ikadub [295]
3 years ago
15

Comment Both propane and benzene are hydrocarbons. As a rule,

Chemistry
1 answer:
kozerog [31]3 years ago
5 0

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

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According to libretexts the answer would be B. decreases.

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In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1
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A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

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We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

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Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

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411g of Ba reacted to produce 138g of Na.

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Therefore, 14.62g of Ba3(PO4)2 is produced.

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